# Difference of two squares

When carrying out mathematical operations it is common to find subtractions between two numbers, however, when finding the subtraction of the squares of two numbers we will say that this is a difference of squares and it is of our particular interest because through the distributive property, we can express it as the product of two factors.

Formally, if $a$ and $b$ are two real numbers, then the difference of their squares will be equal to the sum of the first plus the second, multiplied by the subtraction of the first by the second, that is,

This equality can be deduced by performing the distributive property of the real numbers, let’s see then,

This type of expression is often found in the development of algebraic operations and is used mainly for factoring operations, let’s see in the following examples how to apply this operation

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## Examples

### Example 1

Factorize the expression $5^2 - 3^2$. Note that in this case, we can simply apply the power of each of the summands and perform the subtraction directly.

$5^2 - 3^2 = 25 - 9$

$= 16$

### Example 2

Factorize the expression $x^2 - 9$. We notice that in this case, one of the summands is an x squared and the other one is a nine, so we cannot make the subtraction between them so we apply the difference of squares noting that nine is equal to three squared.

$x^2 - 9 = x^2 - 3^2$

$= (x-3)(x+3)$

### Example 3

Factorize the expression $x^2 - 2$. We notice that in this case, one of the summands is an x-squared and the other is two, so we cannot perform the subtraction between them so we apply the difference of squares noting that two can be rewritten as $2 = \left( \sqrt{2} \right)^2$.

$x^2 - 2 = x^2 -\left( \sqrt{2} \right)^2$

$= \left(x-\sqrt{2} \right) \left(x+\sqrt{2} \right)$

In this way, we can notice that if the square root of a number is not exact, it can be rewritten to use the difference of squares.

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### Example 4

Factorize the expression $8 - x^6$. We notice that in this case, one of the summands is 8 and the other one is x to six, so we cannot make the subtraction between them so we apply the difference of squares noting that eight can be rewritten as $8 = \left( \sqrt{8} \right)^2$ and x to six as $x^6 = \left( x^3 \right)^2$.

$8 - x^6 = \left( \sqrt{8} \right)^2 - \left(x^3 \right)^2$

$= \left(\sqrt{8}-x^3 \right) \left(\sqrt{8}+x^3 \right)$

### Example 5

Factorize the expression $36x^4 - 5x^8$. We notice that in this case, we cannot make the subtraction between them so we apply the difference of squares using the observations exposed in the previous examples.

$36x^4 - 5x^8 = \left( 6x^2 \right)^2 - \left( \sqrt{5}x^4 \right)^2$

$= \left(6x^2-\sqrt{5}x^4 \right) \left(6x^2+\sqrt{5}x^4 \right)$

# The Notable Product

The remarkable product is a particular case of the distributive property that gives us the perfect square trinomial as a result and establishes that, if $a$ and $b$ are two real numbers, the square of the sum of them is equal to the first squared plus twice the product of the first times the second plus the second squared, that is,

This equality can be deduced by performing the distributive property when we multiply the sum of two numbers by that same sum, let’s see then,

Similarly, if $a$ and $b$ are two real numbers, the square of the subtraction between the two is equal to the first squared minus twice the product of the first times the second plus the second squared, that is,

This equality can be deduced by performing the distributive property when we multiply the substraction of two numbers by that same substraction, let’s see then,

This type of expression is often found in the development of algebraic operations because we cannot always carry out the sum that is inside the parentheses, let’s see in the following examples how to apply this operation:

## Ejemplos

### Ejemplo 1

Apply the notable product to expand the expression $(3 + 2)^2$. We add the two elements within the parentheses and square as follows:

$(3 + 2)^2$
$= 5^2$
$= 25$

### Ejemplo 2

Apply the notable product to expand the expression $(3 + \sqrt{2})^2$. Note that one of the addends involved is the square root of two, therefore it cannot be added with three.

$(3 + sqrt{2})^2$
$= 3^2 + 2(3)(sqrt{2}) + (sqrt{2})^2$
$= 9 + 6sqrt{2} + 2$
$= 11+6sqrt{2}$

### Ejemplo 3

Apply the notable product to expand the expression $(\sqrt[3]{6} - 4)^2$. Note that one of the addends involved is the cube root of six, therefore it cannot be subtracted with four.

$(sqrt[3]{6} - 4)^2$
$= (sqrt[3]{6})^2 -2(sqrt[3]{6})(4) + 4^2$
$= (sqrt[3]{6})^2 -8sqrt[3]{6} +16$

### Ejemplo 4

Apply the notable product to expand the expression $(x + 7)^2$. Note that one of the addends involved is an unknown, therefore it cannot be added with seven.

$(x+7)^2$
$= x^2 + 2(x)(7) + 7^2$
$= x^2 +14x + 49$

### Ejemplo 5

Apply the notable product to expand the expression $(2x-8)^2$. Note that one of the addends involved is an unknown multiplied by two, therefore it cannot be subtracted with eight.

$(2x-8)^2$
$= (2x)^2 - 2(2x)(8) + 8^2$
$= 4x^2 - 32x + 64$

### Ejemplo 6

Apply the notable product to expand the expression $(x^2 + x^5)^2$. Note that one of the addends involved is x squared and the other is x raised to five, therefore they cannot be added.

$(x^2 + x^5)^2$
$= (x^2)^2 + 2(x^2)(x^5) + (x^5)^2$
$= x^4 + 2x^7 + x^{10}$

# The Distributive Property

When adding real numbers we have the freedom to associate the numbers involved smoothly, the same happens if we are multiplying real numbers, however, we must be cautious when we come across mixed operations, that is, sums and products at the same time. We will see a property that allows us to operate sums and products at the same time:

The distributive property states that if a number multiplies the sum of two numbers, then the factor involved is distributed among each of the addends. Formally, if $a$, $b$ and $c$ are real numbers, then

We can also apply this property if a subtraction is involved instead of an addition within the parentheses, as follows:

We notice that if we observe this equality from right to left, we are taking the common factor that exists in both addends and we are taking it out to multiply:

$a \cdot b \pm a \cdot c = a \cdot (b \pm c)$

This is one of the most used properties in the calculation of mixed operations and from them, some cases are deduced that facilitate the simplification of mathematical expressions. Let’s see some examples to understand this property well:

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## Examples

### Example 1

Use the distributive property to expand the expression $2 \cdot (1 + 6)$. In this case, it is not necessary to use the distributive property since we can add the numbers that are inside the parentheses and then multiply in the following way:

$2 \cdot (1 + 6) = 2 \cdot 7 = 14$

### Example 2

Use the distributive property to expand the expression $2 \cdot \left (1 + \sqrt {6} \right)$. Note that one of the addends involved is the square root of 6, therefore it cannot be added with 1, so we distribute the factor involved

$2 \cdot \left( 1 + \sqrt{6} \right) = 2 \cdot 1 + 2 \cdot \sqrt{6} = 2 + 2 \sqrt{6}$

### Example 3

Use the distributive property to expand the expression $5 \cdot \left (x - \sqrt {10} \right)$. Note that one of the addends involved is the square root of 10 and the other is an unknown, therefore they cannot be subtracted, so we distribute the factor involved

$5 \cdot \left( x - \sqrt{10} \right) = 5 \cdot x - 5 \cdot \sqrt{10} = 5x - 5\sqrt{10}$

### Example 4

Use the distributive property to expand the expression $x \cdot \left (x + x^2 \right)$. Note that one of the addends involved is an unknown and the other is an unknown squared, therefore they cannot be added, then we distribute the factor involved

$x \cdot \left( x + x^2 \right) = x \cdot x + x \cdot x^2 = x^2 + x^3$

### Example 5

Use the distributive property to take out the common factor of the expression $18 + 3 \sqrt {7}$. Note that $18 = 3 \cdot 6$, then,

$18 + 3\sqrt{7} = 3 \cdot 6 + 3 \sqrt{7} = 3 \cdot \left( 6 + \sqrt{7} \right)$

### Example 6

Use the distributive property to take out the common factor of the expression $x^4 - 8x$. Note that one of the addends involved is an unknown raised to four and the other is 8 times said unknown, therefore they cannot be subtracted, then

$x^4 - 8x = x \cdot x^3 - x \cdot 8 = x \cdot \left( x^3 - 8 \right)$

### Example 7

Use the distributive property to take the common factor of the expression $12x^7 + 15x^4$. These two elements cannot be added, so

$12x^7 + 15x^4 = 3 \cdot 4 \cdot x^4 \cdot x^3 + 3 \cdot 5 \cdot x^4 = 3 x^4 \cdot \left( 4x^3 + 5 \right)$